Răspuns:
Explicație pas cu pas:
[tex]z=\dfrac{1+i}{1-2i} +\dfrac{1+2i}{1-i} =\dfrac{(1+i)(1-i)}{(1-2i)(1-i)}+\dfrac{(1+2i)(1-2i)}{(1-2i)(1-i)}=\dfrac{1^{2}-i^{2}}{(1-2i)(1-i)}+=\dfrac{1^{2}-(2i)^{2}}{(1-2i)(1-i)} =\dfrac{1+1}{1-3i+2i^{2}}+ \dfrac{1+4}{1-3i+2i^{2}}=\dfrac{2}{1-3i-2}+\dfrac{5}{1-3i-2}=-\dfrac{7}{1+3i}=-\dfrac{7(1-3i)}{(1+3i)(1-3i)}=-\dfrac{7(1-3i)}{1^{2}-(3i)^{2}}=-\dfrac{7-21i}{1+9} =-\dfrac{7}{10} +\dfrac{21}{10}i.[/tex]
Deci, Re (z) = -7/10.